Problem 51 - Essential Calculus Early Transcendentals

Question

\(51-60=\) Use logarithmic differentiation or an alternative method to find the derivative of the function. $$y=\left(x^{2}+2\right)^{2}\left(x^{4}+4\right)^{4}$$

Step-by-Step Solution

Answer

The derivative is found using logarithmic differentiation: \( \frac{dy}{dx} = y \left( \frac{4x}{x^2 + 2} + \frac{64x^3}{x^4 + 4} \right) \).

Step 1: Understanding the Function

The function given is \( y = (x^2 + 2)^2 (x^4 + 4)^4 \). It is a product of two sub-functions: \( u(x) = (x^2 + 2)^2 \) and \( v(x) = (x^4 + 4)^4 \). We need to find the derivative of this product.

Step 2: Apply Logarithmic Differentiation

To use logarithmic differentiation, we first take the natural logarithm of both sides: \( \ln y = \ln((x^2 + 2)^2 (x^4 + 4)^4) \). Utilize the logarithmic property: \( \ln(ab) = \ln a + \ln b \). So, \( \ln y = 2\ln(x^2 + 2) + 4\ln(x^4 + 4) \).

Step 3: Differentiate the Logarithmic Form

Differentiate both sides with respect to \( x \). The left side becomes \( \frac{1}{y} \cdot \frac{dy}{dx} \). The right side is differentiated using the chain rule: \( \frac{d}{dx}(2\ln(x^2 + 2)) = \frac{2 \cdot 2x}{x^2 + 2} \) and \( \frac{d}{dx}(4\ln(x^4 + 4)) = \frac{4 \cdot 16x^3}{x^4 + 4} \).

Step 4: Solve for \( \frac{dy}{dx} \)

Combine the differentiated terms: \( \frac{dy}{dx} = y \left( \frac{4x}{x^2 + 2} + \frac{64x^3}{x^4 + 4} \right) \). Substitute back \( y = (x^2 + 2)^2 (x^4 + 4)^4 \) to express \( \frac{dy}{dx} \) solely in terms of \( x \).

Step 5: Simplify the Expression

After substitution, the derivative becomes \( \frac{dy}{dx} = ((x^2 + 2)^2 (x^4 + 4)^4) \left( \frac{4x(x^4 + 4) + 64x^3(x^2 + 2)}{(x^2 + 2)(x^4 + 4)} \right) \). Simplify the expression further by combining like terms in the numerator.

Key Concepts

Derivative

The term "derivative" in calculus refers to the rate at which a function changes as its input changes. In simple terms, it's a way of measuring how a function's output responds to small changes in the input variable. Derivatives are foundational in calculus and are used to find things like velocity (rate of position change) or acceleration (rate of velocity change).

Derivatives are often denoted as \( \frac{dy}{dx} \) to represent the derivative of a function \( y \) with respect to \( x \). This notation tells us how \( y \) changes with a slight change in \( x \). When we apply derivatives, we're often interested in finding specific values or researching how the function behaves at different points.

Product Rule

In calculus, the product rule is a formula used to find the derivative of a product of two functions. The product rule states that if you have two differentiable functions, say \( u(x) \) and \( v(x) \), the derivative of their product is:

• \( (uv)' = u'v + uv' \)

This rule essentially tells us that to derive the product of two functions, we take the derivative of the first function and multiply it by the second, then add the first function multiplied by the derivative of the second.

In our exercise, the function \( y(x) \) was split up into two parts: \( u = (x^2 + 2)^2 \) and \( v = (x^4 + 4)^4 \). We applied the product rule by taking derivatives of \( u \) and \( v \) separately, making the problem simpler and structured.

Chain Rule

The chain rule is a calculus principle used to find the derivative of composite functions. A composite function is a function that is applied within another function, such as \( f(g(x)) \). The chain rule states:

• \( (f \, o \, g)'(x) = f'(g(x)) \cdot g'(x) \)

This means that to differentiate a composite function, you find the derivative of the outer function evaluated at the inner function, and multiply it by the derivative of the inner function.

In our exercise, when differentiating \( (x^2 + 2)^2 \) and \( (x^4 + 4)^4 \), we used the chain rule for each. For example, \( 2(x^2 + 2) \) as the derivative comes from the outer power of 2, and \( 2x \) from the inner function \( x^2 + 2 \), multiplying them to get the complete derivative.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It is fundamentally about change and motion. Two of the main concepts in calculus are differentiation (finding derivatives) and integration (finding integrals or antiderivatives). Both these concepts provide powerful tools for understanding the behavior and properties of functions.

In differentiation, as we did in our exercise, calculus allows us to determine the slope or steepness of a curve at any given point. With the help of rules like the product rule and chain rule, calculus helps unravel complex relationships in a manageable manner. Whether it's predicting trends, optimizing functions, or solving rate problems, calculus stands as a critical tool in the mathematician's toolkit. This exercise illustrates how calculus can be applied to function derivatives using sophisticated techniques to tackle seemingly complicated problems with ease.